CBSE Class 12 Biology 2024: question 5
Set in the CBSE Class 12 board exam of 2024, paper 57/1/1, section A, question 5, for 1 mark. Try it first; the hints and our step-by-step solution are below.
Q5 · 1 mark · MCQCompetency-basedMolecular Basis of Inheritance
You know that there are twenty different types of naturally occurring amino acids and four different types of bases in the DNA. A combination of 3 such bases code for a specific amino acid. If instead there are 96 different amino acids and 12 different bases in the DNA, then the minimum number of combination of bases required to form a codon is :
- (A)6
- (B)8
- (C)2
- (D)4
Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 57/1/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
With 4 bases, triplets give codons.
Hint 2
Find the smallest n for which is at least 96.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
With 12 different bases, a codon of n bases gives combinations. For n = 1 this is 12, which is less than 96. For n = 2 it is 144, which is at least 96.
Therefore the minimum codon length is 2 bases.
Answer: (C) 2
Where the marks are
- Uses to get n = 2 (1 mark)
Revise the chapter
- Chapter: Molecular Basis of Inheritance (notes, formulas, where marks are lost and every past question)
- Topic: Genetic Code and Translation
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